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  Nonlinear potential well (1D Class. Mech)

+ 1 like - 0 dislike

Consider the period $T$ of the bound states of a nonlinear potential well $U(x)$. So $x(t)$ is just bouncing back and forth between $x_0$ and $x_1$.

$T(x_0)\propto\int_{x_0}^{x_1}\frac{\mathrm{d}x}{\sqrt{E-U(x)}}$ where $E=U(x_0)=U(x_1)$.

How do we prove $T$ is differentiable in energy i.e. $\mathrm{d}T/\mathrm{d}x_0$ exists at least for bound states. Leibniz integral rule doesn't really apply here. Do I have to go into the business of action-angle? Thanks.

asked Jul 28, 2015 in General Physics by anonymous [ revision history ]
recategorized Jul 28, 2015

1 Answer

+ 2 like - 0 dislike

Write $f(x_0,E,\delta)$ for the integral, taken between $x_0+\delta$ and $x_1-\delta$, where $\delta>0$ is tiny. Then you may differentiate under the integral with respect to E to get $\frac{df}{dE}(x_0,E,\delta)$. Then you must show that the limit of the result remains well-defined when $\delta\to 0$ to get $\frac{df}{dE}(x_0,E,0)$. Now you may use the chain rule to calculate the derivative of $T(x_0)=f(x_0,U(x_0),0)$.

answered Jul 29, 2015 by Arnold Neumaier (15,787 points) [ revision history ]

So basically you treat $E$ as an independent variable in the first place, evaluate the integral, then plug in $E=U(x_0)$ and differentiate in $x_0$ via chain rule. Sounds brilliant. Thanks.

I was thinking of transforming the integration domain of any $T(x_0)$ to a fixed interval say $(0,\frac{1}{2})$ via change of variable. Then show that the derivative of the new integrand with respect to $x_0$ exists a.e and integrable.

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